please help how do i go about solving this?? undefinedundefinedundefined A jogger runs in a straight line with an average velocity of 4:7 m/s for 5.1 min, and then with an average velocity of 4.1 m/s for 3 min. What is her total displacement? Answer in units of meters
Well, it all depends what the direction is. If he runs in the same direction both times, you have to add the 2 displacements to find the total displacement.
Even if the runs in the opposite direction on the second time interval,the TOTAL distance would still be gotten by adding the 2. Daniel.
Aren't the words "distance travelled" and "displacement" different. I though displacement meant distance and direction from the original starting point.
for the same question, to get the avg. velocity during the time what should i do? i know it is not just to add the 2 velocities given and divide by 2
I think that u've given yourself the answer. Besides,since in the problem we're not given the directions (so that we would think it as a vector projection problem),it's natural to assume the simpler version,which is what the teacher would want,right...? :tongue2: Daniel.
new question A 500-kilogram sports car accelerates uni- formly from rest, reaching a speed of 30 me- ters per second in 6 seconds. During the 6 seconds, the car has traveled a distance of how many meters?? i got 180m is that correct? i used : distance = avg. velocity * change in time = (30m.s) * 6 = 180m
Careful... 30m/s is not the average velocity, but rather the final velocity. Remember, average velocity is total displacement divided by the total time, or in this case (if a is constant), [tex]\frac{V_i + V_f}{2} = \frac{0+30m/s}{2}[/tex]
Isn't the average the sum of the 2 velocities divided by 2? In this case, the displacement would be: [tex]d = V_{ave}t[/tex]
I's not "side talking". I'm referring to the "new question". Here, average velocity IS involved in finding the distance. Anyways, gotta go. I hope I didn't irritate you too much Daniel